Projectile Motion: Launching Up a Hill

[Shackleford]

http://i111.photobucket.com/albums/n149/camarolt4z28/14q.jpg?t=1284229871

Here is my work. My denominator is different than the book's answer for (a). I still don't know how to maximize alpha for (b) or maximize the range for (c).

http://i111.photobucket.com/albums/n149/camarolt4z28/141.jpg?t=1284229758
http://i111.photobucket.com/albums/n149/camarolt4z28/142.jpg?t=1284229758

http://i111.photobucket.com/albums/n149/camarolt4z28/14a.jpg?t=1284229759

 

[Mindscrape]

If you're going to use the coordinate that you set, you'll have to watch out for correctly accounting for gravity. In both coordinate frames you will have a contribution from gravity. You actually even have your initial drawing of the coordinate system labeled incorrectly.

 

[Shackleford]

If you're going to use the coordinate that you set, you'll have to watch out for correctly accounting for gravity. In both coordinate frames you will have a contribution from gravity. You actually even have your initial drawing of the coordinate system labeled incorrectly.

I thought I did correctly account for gravity. Determining the effective gravity in this case is akin to dealing with a block on a plane.

What's wrong with the coordinate system? My understanding of the problem is a projectile with elevation alpha being shot up a hill of elevation. I suppose another interpretation could be a projectile shot with elevation alpha with respect to a hill of elevation beta.

 

[rl.bhat]

Select the co-ordinate system such that the direction OX i.e. X-axis along the surface of the inclined plane and Y-axis perpendicular to the surface.

Resolve vo into two components.
vox = vo*cos(α - β)
voy= vo*sin(α - β)
gx = g*sinβ and gy = g*cosβ

Τhe equation of the projectile motion is given by
y = vo*sin(α - β)*T - 1/2*g*cosβ*Τ^2

The time of flight T can be found by putting y = 0.

T = 2vo*sin(α - β)/g*cosβ

The horizontal distance x is given by
x = vo*cosα*T = 2vo^2*sin(α - β)*cosα/g*cosβ
Range d = x/cosβ = 2vo^2*sin(α - β)*cosα/g*cos^2(β)

Using the trigonometric identity 2sin(A)con(B) = sin(A+B) + sin(A-B), You can write

d = [vo^2/g*cos^2(β)]*[sin(2α - β) + sin(β)]

To find the angle of projection for maximum range,find derivative of d with respect to α and equate it to zero.

Substitute this angle in d to get the maximum range.

 

[Shackleford]

Wait a second. I'm not sure why I have it backwards. Let me re-work it.

 

[Shackleford]

I found x to be

x' = -(1/2) t^2 sin β + v0 cos (α - β)t.

I thought they wanted the x distance up the hill, or x', not the "original" x. That of course is simply x = v0 cos (α) t since there is no effective gravitational force in that direction. In x' coordinate transformation, there is an effective gravitational force along x' and y'.

Okay, I see what you did. To find d up the hill, you compared the x distance in the "original" system with d up the hill by cosine of beta determined by time T with y' = 0.

Well, from my d(d)/dα, I end up with cos(α - β) = 0.

d(d)/dα = [vo^2/g*cos^2(β)]*[2 cos(2α - β) + 0]

cos(2α - β) = 0

2α - β = n*(pi/2), where is n odd integer

α = β/2 + pi/4

I plugged in that angle, but I didn't get dmax.

 

[rl.bhat]

Range d = x/cosβ = 2vo^2*sin(α - β)*cosα/g*cos^2(β)

= 2vo^2*sin[π/4 +β/2 - β]*cos(π/4 +β/2)/g*(1 - sin^2β)

= 2vo^2*sin[π/4 - β/2]*cos(π/4 +β/2)/g*(1 - sin^2β)

= vo^2*[sinπ/2 - sinβ]/g*(1+sinβ)(1-sinβ)

= vo^2*/g*(1 + sinβ)

 

[Mindscrape]

I personally think this problem would be easier with a coordinate system in line the legs of the triangle, or the hill. I guess whatever is easiest, though you seemed to get in trouble with your gravitation earlier.

Of course, in the end you will get what rl.bhat posted.

Your condition is right. You should get

$$d_{max} = \frac{v_0^2}{g(1+sin \beta)}$$

which makes sense because you know that it will go furthest when pointed directly up the hill, when beta is zero.

 

[Shackleford]

I personally think this problem would be easier with a coordinate system in line the legs of the triangle, or the hill. I guess whatever is easiest, though you seemed to get in trouble with your gravitation earlier.

Of course, in the end you will get what rl.bhat posted.

Your condition is right. You should get

$$d_{max} = \frac{v_0^2}{g(1+sin \beta)}$$

which makes sense because you know that it will go furthest when pointed directly up the hill, when beta is zero.

I completely overlooked/forgot about the effective gravitation in the x' direction and switched my sines and cosines.

Initially, I plugged in the alpha-max in the last distance equation after the trigonometric identity was inserted. That of course yielded something different.

 

[rl.bhat]

I personally think this problem would be easier with a coordinate system in line the legs of the triangle, or the hill. I guess whatever is easiest, though you seemed to get in trouble with your gravitation earlier.

Of course, in the end you will get what rl.bhat posted.

Your condition is right. You should get

$$d_{max} = \frac{v_0^2}{g(1+sin \beta)}$$

which makes sense because you know that it will go furthest when pointed directly up the hill, when beta is zero.

Here is the simple method. Equation of the projectile motion is

y = x*tanα - 1/2*g*x^2/v^2cos^2α

Equation of the hill is given by

y = tanβ*x. Therefore

x*taνβ = x*tanα - 1/2*g*x^2/v^2cos^2α

So x = (tanα - tanβ)*2v^2*cos^2α/g = 2v^2cosα*sin(α - β)/g*cosβ.

And d = x/cosβ